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The line $A B$ cuts off equal intercepts $2 a$ from the axes. From any point $P$ on the line $A B$ perpendiculars $P R$ and $P S$ are drawn on the axes. Locus of mid-point of $R S$ is
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Verified Answer
The correct answer is:
$x+y=a$
$\Rightarrow$
$$
\begin{array}{l}
\frac{x}{2 a}+\frac{y}{2 a}=1 \\
x+y=2 a
\end{array}
$$
Let the coordinates of the mid-point of $\mathrm{RS}$ be $(\mathrm{h}, \mathrm{k})$
So. $R$ and $S$ are $(2 h, 0)$ and $(0,2 k)$ Hence, the mid-point of $R S$ is $(h, k)$
P lies on line AB
Then. from Eq we have
$$
2 h+2 k=2 a
$$
$\Rightarrow \quad h+k=a$
So, the locus of $(h, k)$ is $x+y=a$.
$$
\begin{array}{l}
\frac{x}{2 a}+\frac{y}{2 a}=1 \\
x+y=2 a
\end{array}
$$
Let the coordinates of the mid-point of $\mathrm{RS}$ be $(\mathrm{h}, \mathrm{k})$
So. $R$ and $S$ are $(2 h, 0)$ and $(0,2 k)$ Hence, the mid-point of $R S$ is $(h, k)$
P lies on line AB
Then. from Eq we have
$$
2 h+2 k=2 a
$$
$\Rightarrow \quad h+k=a$
So, the locus of $(h, k)$ is $x+y=a$.
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